Question: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}-2x-8y &= 1 \\ -x+6y &= 3\end{align*}$
Solution: Begin by moving the $x$ -term in the second equation to the right side of the equation. $6y = x+3$ Divide both sides by $6$ to isolate $y$ $y = {\dfrac{1}{6}x + \dfrac{1}{2}}$ Substitute this expression for $y$ in the first equation. $-2x-8({\dfrac{1}{6}x + \dfrac{1}{2}}) = 1$ $-2x - \dfrac{4}{3}x - 4 = 1$ Simplify by combining terms, then solve for $x$ $-\dfrac{10}{3}x - 4 = 1$ $-\dfrac{10}{3}x = 5$ $x = -\dfrac{3}{2}$ Substitute $-\dfrac{3}{2}$ for $x$ back into the top equation. $-2( -\dfrac{3}{2})-8y = 1$ $3-8y = 1$ $-8y = -2$ $y = \dfrac{1}{4}$ The solution is $\enspace x = -\dfrac{3}{2}, \enspace y = \dfrac{1}{4}$.